∗
f(px + (1 − p)y
)+ f
((1 − p)x + py
)= f(x) + f(y), (x, y ∈ I)
0 < p < 1 f : I → R
p
0 < p < 1
f(px + (1− p)y
)+ f
((1− p)x + py
)= f(x) + f(y),
∗
p
©
f : I → R I ⊂ Rx, y ∈ I Sp(I) f : I → R
Sp(I) = S1−p(I)
f ∈ Sp(I) k
Ak : Rk → R k = 0, 1, 2
A2
(px, (1− p)x
)= 0 (x ∈ R)
f(x) = A2(x, x) + A1(x) + A0
x ∈ I
pp ∈ ]0, 1[ f : I → R p I
f(px + (1− p)y
)+ f
((1− p)x + py
)� f(x) + f(y)
x, y ∈ I f : I → R −f : I → R pI f ∈ Sp(I)
p I
f ∈ Sp(I) ξ ∈ Iδ > 0 ]ξ − δ, ξ + δ[ ⊂ I f | ]ξ − δ, ξ + δ[
Sp
(]ξ − δ, ξ + δ[
)f : I → R ξ ∈ I
I ξ f fI
0 < p < 1A2
p
f(αx + (1− α)y
)+ f
((1− α)x + αy
)= f
(βx + (1− β)y
)+ f
((1− β)x + βy
)x, y ∈ I f : I → R 0 < α < 1 0 < β < 1
α �∈ {β, 1− β}
p
A2 : R2 → Rα
A2(αx, x) = 0 x ∈ R,
−α α−α
α α∑n
i=1 tiαi = 0 ti ∈ Z i = 0, . . . , n
tn �= 0
A2(αkx, y) = (−1)kA2(x, αky)
k � 0 x, y ∈ R k = 0
0 = A2
(α(x− y), x− y
)= A2(αx, x)−A2(αy, x)−A2(αx, y) + A2(αy, y)
= −A2(αy, x)−A2(αx, y)
x, y ∈ R A2(αx, y) = (−1)A2(x, αy)k = 1 k � 1
A2(αk+1x, y) = A2(ααkx, y) = −A2(αkx, αy)
= (−1)(−1)kA2(x, αkαy) = (−1)k+1A2(x, αk+1y),
k + 1
β :=n∑
i=0
ti(−α)i �= 0.
x, y ∈ R z :=y
β
0 = A2(0, z) = A2
(( n∑i=0
tiαi
)x, z
)=
n∑i=0
tiA2(αix, z)
=n∑
i=0
ti(−1)iA2(x, αiz) = A2
(x,
( n∑i=0
ti(−1)iαi
)z
)= A2(x, βz) = A2(x, y),
A2 �
α −αα
A2 : R2 → R α α−α α
α βa : R → R
a(αx) = βa(x) x ∈ R.
α βα a : R → R
a(αx) = βa(x) x ∈ R.
α α −αa : R → R
a(αx) = −αa(x) x ∈ R.
a : R → R
A2(x, y) := a(x)y + a(y)x x, y ∈ R.
A2 : R2 → R
A2(αx, x) = a(αx)x + a(x)αx = −αa(x)x + a(x)αx = 0
x ∈ R �
0 < p < 1 f ∈ Sp(I)
2f
(x + y
2
)= f(x) + f(y) (x, y ∈ I)?
0 < p < 11−p
p −1−pp
f ∈ Sp(I)1−p
p1−p
p
−1−pp f ∈ Sp(I)
A2 : R2 → Rpx = y
A2
(1− p
py, y
)= 0 y ∈ R
A21−p
p −1−pp
A2 : R2 → R
A2
(1− p
px, x
)= 0 x ∈ R,
x �→ A(x, x) x ∈ I
�
pp
p pp p
p 1−pp −1−p
p
p ∈ ]0, 1[ p �= 12
1−pp −1−p
p
p p2p−1
α β
0 < α < β � 12.
a < b a, b ∈ I Pa,b(I)(a, a)
(αa + (1− α)b, (1− α)a
+ αb) (
(1− α)a + αb, αa + (1− α)b)
(b, b)
P (I) :=⋃
a,b∈I, a<b
Pa,b(I).
f : I → Rx, y ∈ I p :=
α+β−12α−1 ∈ ]0, 1[ f : I → R
f(pu + (1− p)v
)+ f
((1− p)u + pv
)= f(u) + f(v)
(u, v) ∈ P (I) ⊂ I2 P (I) R2
{u = αx + (1− α)y
v = (1− α)x + αy(x, y) ∈ I2,
(x, y) ∈ I2 (u, v) ∈ I2
{(u, v) =(αx + (1− α)y, (1− α)x + αy
)| x, y ∈ I} = P (I),
P (I)x y ⎧⎪⎪⎨⎪⎪⎩
x =α
2α− 1u +
α− 12α− 1
v
y =α− 12α− 1
u +α
2α− 1v
(u, v) ∈ P (I).
f(u) + f(v)
= f
(β
(α
2α− 1u +
α− 12α− 1
v
)+ (1− β)
(α− 12α− 1
u +α
2α− 1v
))+ f
((1− β)
(α
2α− 1u +
α− 12α− 1
v
)+ β
(α− 12α− 1
u +α
2α− 1v
))= f
(pu + (1− p)v
)+ f
((1− p)u + pv
),
�
f : I → Rx, y ∈ I f ∈ Sp(I) p := α+β−1
2α−1 ∈ ]0, 1[
(u, v) ∈ P (I) ⊂ I2
ξ ∈ I δ > 0]ξ − δ, ξ + δ[ ⊂ I f | ]ξ − δ, ξ + δ[ ∈ Sp
(]ξ − δ, ξ + δ[
)I2 :=
{(ξ, ξ) | ξ ∈ I
}P (I)
δ > 0 ξ ∈ I ]ξ − δ, ξ + δ[2 ⊂ P (I)f | ]ξ − δ, ξ + δ[ ∈ Sp
(]ξ − δ, ξ + δ[
)�
α, β ∈ ]0, 1[ α �∈ {β, 1− β} A2 : R2 → R
A2
(αx, (1− α)x
)= A2
(βx, (1− β)x
)(x ∈ R)
A2
(α + β − 1
α− βx, x
)= 0 (x ∈ R).
A2
((α + β − 1)y, (α− β)y
)= A2
(βy, (1− β)y
)−A2
(αy, (1− α)y
)y ∈ R
A2
((α + β − 1)y, (α− β)y
)= 0 (y ∈ R).
x = (α− β)y�
α,β ∈ ]0, 1[ α �∈ {β,1−β} f : I →R
f(αx + (1− α)y
)+ f
((1− α)x + αy
)= f
(βx + (1− β)y
)+ f
((1− β)x + βy
)(x, y ∈ I) k Ak : Rk
→ R (k = 0, 1, 2)
A2
(αx, (1− α)x
)= A2
(βx, (1− β)x
)(x ∈ R)
f(x) = A2(x, x) + A1(x) + A0 x ∈ I.
f : I → Rf ∈ Sp(I) p = α+β−1
2α−1 ∈ ]0, 1[k Ak : Rk
→ R (k = 0, 1, 2)
A2
(px, (1− p)x
)= 0 (x ∈ R)
A2
(p
1− py, y
)= 0 (y ∈ R),
A2
(α + β − 1
2α− 1· 2α− 1
α− βy, y
)= 0 (y ∈ R).
�f
x �→ A2(x, x) x ∈ I
α, β ∈ ]0, 1[ α �∈ {β, 1−β}A2 : R2 →R
α+β−1α−β −α+β−1
α−β
�
x �→ A2(x, x) x ∈ I
t :=α + β − 1
α− β− t t.
A(α, β)A2 : R2 → R α, β ∈ ]0, 1[ α �∈ {β, 1− β}
α β12
α+β−1α−β A(α, β) �= ∅
α β 12 A(α,β) = ∅
a ∈ {2√
2, 2 3√
2}
α := limn→∞
(1− 1
n
)n
=1e, β :=
a
a− 1− a + 1
a− 11e,
α, β ∈ ]0, 1[ α �∈ {β, 1− β} α+β−1β−α = 1
a A(α, β) �= ∅a = 2
√2 A(α, β) = ∅ a = 2 3
√2
t